<u>Answer:</u> The molarity of calcium hydroxide in the solution is 0.1 M
<u>Explanation:</u>
To calculate the concentration of base, we use the equation given by neutralization reaction:

where,
are the n-factor, molarity and volume of acid which is 
are the n-factor, molarity and volume of base which is 
We are given:

Putting values in above equation, we get:

Hence, the molarity of
in the solution is 0.1 M.
Answer:
a. 1.12 L
Explanation:
Step 1: Write the balanced equation for the photosynthesis
6 CO₂(g) + 6 H₂O(l) ⇒ C₆H₁₂O₆(s) + 6 O₂(g)
Step 2: Calculate the moles corresponding to 2.20 g of CO₂
The molar mass of CO₂ is 44.01 g/mol.
2.20 g × 1 mol/44.01 g = 0.0500 mol
Step 3: Calculate the moles of O₂ produced
The molar ratio of CO₂ to O₂ is 6:6. The moles of O₂ produced are 6/6 × 0.0500 mol = 0.0500 mol
Step 4: Calculate the volume occupied by 0.0500 moles of O₂ at STP
At STP, 1 mole of O₂ occupies 22.4 L.
0.0500 mol × 22.4 L/1 mol = 1.12 L
Answer;
The partial negative charge on oxygen would stick out less and be less able to participate in hydrogen bonding.
Explanation;
Water is a polar molecule because the electrons are not shared equally, they're closer to the oxygen atom than the hydrogen.
-Normally, the water molecule is a bent shape because of the pair of lone electrons - they repulse each other and exert a compression to the hydrogen atoms at a slight 104º angle. It is a bent molecular geometry that results from tetrahedral electron pair geometry.
-The 2 lone electron pairs exerts a little extra repulsion on the two bonding hydrogen atoms to create a slight compression to a 104 degrees bond angle. Therefore, the water molecule is bent molecular geometry because the lone electron pairs.
Thus, If water were a linear molecule like co2, electrostatic interactions between water molecules would be much weaker, then the partial negative charge on oxygen would stick out less and be less able to participate in hydrogen bonding.
Answer: 
Explanation:

cM 0 0
So dissociation constant will be:

Given: c = 0.15 M
pH = 1.86
= ?
Putting in the values we get:
Also ![pH=-log[H^+]](https://tex.z-dn.net/?f=pH%3D-log%5BH%5E%2B%5D)
![1.86=-log[H^+]](https://tex.z-dn.net/?f=1.86%3D-log%5BH%5E%2B%5D)
![[H^+]=0.01](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D0.01)
![[H^+]=c\times \alpha](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3Dc%5Ctimes%20%5Calpha)


As ![[H^+]=[ClCH_2COO^-]=0.01](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D%5BClCH_2COO%5E-%5D%3D0.01)

![K_a=1.67\times 10^{-3]](https://tex.z-dn.net/?f=K_a%3D1.67%5Ctimes%2010%5E%7B-3%5D)
Thus the vale of
for the acid is 